A solution containing a period-IV cation gives a precipitate on passing H$$_2$$S. A solution of this precipitate in dil. HCl produces a white precipitate with NaOH solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is:

First we recall the principle of qualitative analysis: cations of the fourth analytical group are those whose sulphides are precipitated when colourless hydrogen sulphide gas is passed through the original solution kept slightly alkaline with ammonium chloride and ammonium hydroxide. The common period-IV cations that behave in this way are $$Mn^{2+},\; Fe^{2+},\; Co^{2+},\; Ni^{2+},\; Zn^{2+}.$$ In the present question the list of possibilities has already been narrowed down to four cations of this group.

We are told that on passing $$H_2S$$ a precipitate appears. Let us represent the general reaction that takes place when a divalent metal ion $$M^{2+}$$ of this group encounters hydrogen sulphide in alkaline medium:

$$M^{2+}+ S^{2-}\;\rightarrow\; MS\downarrow$$

Here $$MS$$ is the metal sulphide. Among the four given cations the colours of these sulphides are well known:

$$ZnS$$ is white, $$MnS$$ is flesh-coloured (light buff), $$NiS$$ is black and $$CoS$$ is black.

The statement of the problem merely says “a precipitate” without giving its colour, so at this stage all four contenders are still formally possible.

The precipitate is next treated with dilute hydrochloric acid and it dissolves. We write the generic dissolution reaction with zinc as an example:

$$ZnS + 2\,HCl \;\rightarrow\; ZnCl_2 + H_2S\uparrow$$

$$ZnCl_2$$ thus formed is a colourless solution containing $$Zn^{2+}$$ ions. Importantly, both $$NiS$$ and $$CoS$$ are insoluble in dilute hydrochloric acid, whereas $$ZnS$$ and $$MnS$$ dissolve. Hence, the very fact that the sulphide dissolves rules out $$Ni^{2+}$$ and $$Co^{2+}$$ from further consideration.

Now we have to distinguish between the two remaining possibilities, $$Mn^{2+}$$ and $$Zn^{2+}.$$ For that we look at the next operation: the acid solution gives a white precipitate with sodium hydroxide solution. Let us write the hydroxide formation reactions for the two ions still in the race:

For zinc:

$$ZnCl_2 + 2\,NaOH \;\rightarrow\; Zn(OH)_2\downarrow \;+\; 2\,NaCl$$

For manganese:

$$MnCl_2 + 2\,NaOH \;\rightarrow\; Mn(OH)_2\downarrow \;+\; 2\,NaCl$$

Although freshly precipitated $$Mn(OH)_2$$ is almost white, it very rapidly absorbs oxygen from the air and turns brown due to the formation of hydrated $$MnO_2$$. In contrast, $$Zn(OH)_2$$ is a persistent white gelatinous precipitate. The question explicitly says “white precipitate”, which is fully consistent with $$Zn(OH)_2$$ and not with the easily darkening $$Mn(OH)_2$$. This observation strongly favours $$Zn^{2+}$$.

The decisive confirmation comes from the final test: the solution gives a bluish-white precipitate with basic potassium ferrocyanide, $$K_4[Fe(CN)_6]$$. We first state the qualitative formula that governs such tests:

$$n\,M^{2+} + [Fe(CN)_6]^{4-} \;\rightarrow\; M_n[Fe(CN)_6]\downarrow$$

For zinc the reaction is written explicitly as

$$2\,Zn^{2+} + [Fe(CN)_6]^{4-} \;\rightarrow\; Zn_2[Fe(CN)_6]\downarrow$$

The precipitate $$Zn_2[Fe(CN)_6]$$ is described in the literature as bluish white. In contrast, $$Mn^{2+}$$ with the same reagent gives a light pink or flesh-coloured precipitate, quite distinct from bluish white. Therefore the observation provided in the problem matches $$Zn^{2+}$$ and no other.

Let us collect the evidence:

1. Sulphide precipitated by $$H_2S$$ dissolves in dilute $$HCl$$   →   eliminates $$Ni^{2+}$$ and $$Co^{2+}$$.
2. Hydroxide obtained with $$NaOH$$ remains white   →   favours $$Zn^{2+}$$ over $$Mn^{2+}$$.
3. Basic ferrocyanide gives bluish-white precipitate   →   characteristic of $$Zn^{2+}$$.

All three experimental facts converge on the same conclusion:

$$\boxed{Zn^{2+}}$$

Hence, the correct answer is Option B.